# Gibbs Free Energy Calculation

1. Jul 24, 2015

### sgstudent

ΔG=ΔH-TΔSsystem and ΔSsystem=qp/T and isn't qp=ΔH? Which means ΔG=0 always which is obviously wrong.

But the derivation for Gibbs Free Energy is based on qsurrounding=-qsystem so shouldn't the term be equals to 0?

2. Jul 24, 2015

### Ygggdrasil

ΔSsystem=qp/T is true only for reversible processes. What do you know about the ΔG of a reversible process?

3. Jul 25, 2015

### sgstudent

It is 0. So for a reversible process qp=ΔH and so ΔG=0?

For an irreversible process how would the ΔS look like?

Thanks

4. Aug 7, 2015

### Staff: Mentor

Are you aware that ΔG and ΔS are functions only of the initial and final thermodynamic equilibrium states of a system, and are independent of path between the two states, whether reversible or irreversible? From what you have learned, do you know how to determine the change in S between two thermodynamic equilibrium states of a system? Are you familiar with the equation dG = -SdT + VdP, which applies to a differential change in temperature and pressure between two closely neighboring thermodynamic equilibrium states of a system?

I think your real question is "how do one determine ΔG and ΔS between two thermodynamic equilibrium states of a system."

Chet

5. Aug 14, 2015